1. Define the function $f(n) = \frac{\sigma(n)}{n}$, where $\sigma(n)$ is the sum of the natural divisors of $n$.
2. We need to show that there are infinitely many integers $m$ such that $\forall k \in \{1, 2, \dots, m-1\}$, $\frac{\sigma(m)}{m} > \frac{\sigma(k)}{k}$. These integers $m$ are called superabundant numbers.
3. Consider the product $\prod_{i=1}^\infty \left(1 + \frac{1}{p_i}\right)$, where $p_i$ denotes the $i$-th prime number.
4. We claim that this product diverges. To see why, note that:
$$\prod_{i=1}^\infty \left(1 + \frac{1}{p_i}\right) \geq 1 + \frac{1}{p_1} + \frac{1}{p_2} + \frac{1}{p_3} + \cdots$$
5. It is a well-known result that the sum of the reciprocals of the primes diverges. Specifically, the series $\sum_{i=1}^\infty \frac{1}{p_i}$ diverges.
6. Since the product $\prod_{i=1}^\infty \left(1 + \frac{1}{p_i}\right)$ diverges, the function $f(n)$ can get arbitrarily large.
7. To see why $f(n)$ can get arbitrarily large, consider the values of $f$ at products of distinct primes:
$$f(p_1) = 1 + \frac{1}{p_1}, \quad f(p_1 p_2) = \left(1 + \frac{1}{p_1}\right)\left(1 + \frac{1}{p_2}\right), \quad f(p_1 p_2 p_3) = \left(1 + \frac{1}{p_1}\right)\left(1 + \frac{1}{p_2}\right)\left(1 + \frac{1}{p_3}\right), \quad \text{and so on.}$$
8. Since the product $\prod_{i=1}^\infty \left(1 + \frac{1}{p_i}\right)$ diverges, $f(n)$ can become arbitrarily large.
9. Because $f(n)$ can become arbitrarily large, it must achieve a new maximum infinitely often. Each time $f(n)$ achieves a new maximum, the corresponding $n$ is a superabundant number.

$\blacksquare$